Difference between revisions of "2009 AMC 8 Problems/Problem 3"
(10 intermediate revisions by 8 users not shown) | |||
Line 25: | Line 25: | ||
label(rotate(90)*"MILES", (-5,11), W);</asy> | label(rotate(90)*"MILES", (-5,11), W);</asy> | ||
− | <math> \textbf{(A)}5\qquad\textbf{(B)}5.5\qquad\textbf{(C)}6\qquad\textbf{(D)}6.5\qquad\textbf{(E)}7 </math> | + | <math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ 5.5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 6.5\qquad\textbf{(E)}\ 7 </math> |
+ | |||
+ | ==Solution== | ||
+ | Suzanna's speed is <math>\frac{1}{5}</math>. This means she runs <math>\frac{1}{5} \cdot 30 = \boxed{ \textbf{(C) }6 }</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | From the graph, we can see that every <math>5</math> minutes Suzanna goes, her distance increases by <math>1</math>. Since half an hour is <math>10</math> minutes away, she would go <math>2</math> more miles. <math>4+2</math> is <math>6</math>, so the answer is <math>\boxed{ \textbf{(C) }6 }</math> | ||
+ | |||
+ | ~Trex226 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/USVVURBLaAc?t=117 | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/agfRvXTPxVc | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=2|num-a=4}} | {{AMC8 box|year=2009|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Latest revision as of 08:22, 30 May 2023
Problem
The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of a half an hour at the same speed, how many miles would she have ridden?
Solution
Suzanna's speed is . This means she runs
Solution 2
From the graph, we can see that every minutes Suzanna goes, her distance increases by
. Since half an hour is
minutes away, she would go
more miles.
is
, so the answer is
~Trex226
Video Solution
https://youtu.be/USVVURBLaAc?t=117
Video Solution 2
~savannahsolver
See Also
2009 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.